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Found in: Page 292

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# A chain of length L and total mass M is released from rest with its lower end just touching the top of a table as shown in Figure P9.96a. Find the force exerted by the table on the chain after the chain has fallen through a distance x as shown in Figure P9.96b. (Assume each link comes to rest the instant it reaches the table.

The total force exerted by the table on chain during falling distance x is $\frac{3Mgx}{L}$.

See the step by step solution

## Step 1: Identification of given data

The mass of chain is $M$

The length of chain is $L$

The distance fallen by the chain through hole is $x$

## Step 2: Linear Mass Density

The linear mass density of an object is the mass of the unit length of the object. it is given as –

${\mathbit{\lambda }}{\mathbf{=}}\frac{\mathbf{M}}{\mathbf{L}}$

## Step 3: Determination of force exerted by table on the chain

The mass of small element of chain is given as:

$dm=\lambda ·dx\phantom{\rule{0ex}{0ex}}=\left(\frac{M}{L}\right)dx$

The weight of chain length fallen is given as:

$dw=\lambda gx\phantom{\rule{0ex}{0ex}}=\left(\frac{M}{L}\right)gx$

The speed of the chain after falling distance x is given as:

$v=\sqrt{2gx}$

The force exerted on chain after falling distance x is given as:

$F=v\left(\frac{dm}{dt}\right)$

For the given values, the above equation becomes-

$F=v\left(\frac{\left(\left(\frac{M}{L}\right)dx\right)}{dt}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{M}{L}\right)v\left(\frac{dx}{dt}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{M}{L}\right)v\left(v\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{M}{L}\right){v}^{2}$

solving further,

$F=\left(\frac{M}{L}\right){\left(\sqrt{2gx}\right)}^{2}\phantom{\rule{0ex}{0ex}}F=\frac{2Mgx}{L}$

The total force exerted by the table on chain during falling distance x is given as:

${F}_{t}=F+w$

For the given values, the above equation becomes-

${F}_{t}=\left(\frac{2Mgx}{L}\right)+\left(\frac{Mgx}{L}\right)\phantom{\rule{0ex}{0ex}}=\frac{3Mgx}{L}$

Therefore, the total force exerted by table on chain during falling distance x is $\frac{3Mgx}{L}$ .